Proving that nothing does not exist

charlie_sheep

I applied some mathematical view to the daily language while studying demonstration.

Proving that nothing does not exist

Consider the following hypothesis by definition:
1. (There's nothing) -> (There's the absence of everything)
2. (There's nothing) -> (There's the absence of everything) -> (There's the absence of the absence of everything)¹ -> (There's everything) -> ~(There's the absence of everything)
And consider the following hypothesis by logic:
3. (There's nothing) v ~(There's nothing)²
4. ~[(There's the absence of everything) ^ ~(There's the absence of everything)]³
By 1 and 2, we have:
5. (There's nothing) -> (There's the absence of everything) ^ ~(There's the absence of everything)
By 5 and 3, we have:
6. [(There's the absence of everything) ^ ~(There's the absence of everything)] v ~(There's nothing)
By 6 and 4, we have:
7. ~(There's nothing)
Q.E.D.

¹ - Cause "everything" includes the "absence of everything", since "absence of everything" is something.
² - Law of excluded middle
³ - Law of non-contradiction

As a result, the space is not full of nothing. Cause nothing does not exist.

Is the demonstration right?

HallsofIvy

Was this a joke? Because all you have is one long play on words.

charlie_sheep

No, it's not a joke.

Long play on words you say, so let me take off the words:

Let A and B be propositions.

Proving ~A

Consider the following hypothesis by definition:
1. A -> B
2. A -> ~B
And consider the following hypothesis by logic:
3. (A v ~A)²
4. ~(B ^ ~B)³
By 1 and 2, we have:
5. A -> (B ^ ~B)
By 5 and 3, we have:
6. (B ^ ~B) v ~A
By 6 and 4, we have:
7. ~A
Q.E.D.

² - Law of excluded middle
³ - Law of non-contradiction

Mute

From these two lines it follows that B and ~B, giving you a contradiction. Given a contradiction, you can derive any conclusion.

micromass

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